Question

A 60 W load is connected to the secondary of an ideal transformer whose primary draws line voltage. If a current

of 0.54 A flows in the load, the current in the primary coil is

• A. 0.27 mA
• B. 2.7 Ma
• C. 0.27 A
• D. 10 A

Answer 1

Answer: (C)

0.27 A

Answer 2

: Here,

$P_{s} = 60W,I_{s} = 0.54A,V_{P} = 220V$

$\therefore V_{s} = \frac{P_{s}}{I_{s}} = \frac{60}{0.54} = 111V$

As $\frac{V_{s}}{V_{p}} = \frac{I_{p}}{I_{s}},$ for an ideal transformer

$\therefore I_{P} = \frac{V_{s}}{V_{P}} \times I_{s} = \frac{111}{220} \times 0.54 = 0.27A$